Question: Find the number of different complex numbers $z$ with the properties that $|z|=1$ and $z^{6!}-z^{5!}$ is a real number.
Solution: Since $|z| = 1,$ we can write $z = \operatorname{cis} \theta,$ where $0^\circ \le \theta < 360^\circ.$  Then
\[z^{6!} - z^{5!} = \operatorname{cis} (720 \theta) - \operatorname{cis} (120 \theta)\]is a real number.  In other words, $\sin 720 \theta - \sin 120 \theta = 0.$  From the sum-to-product formulas,
\[2 \cos 420 \theta \sin 300 \theta = 0.\]If $\cos 420 \theta = 0,$ then $420 \theta$ must be an odd multiple of $90^\circ,$ i.e.
\[420 \theta = (2n + 1) 90^\circ\]for some integer $n.$  The possible values of $n$ are 0, 1, 2, $\dots,$ 839, for 840 solutions.

If $\sin 300 \theta = 0,$ then $300 \theta$ must be a multiple of $180^\circ,$ i.e.
\[300 \theta = m \cdot 180^\circ\]for some integer $m.$  The possible values of $m$ are 0, 1, 2, $\dots,$ 599, for 600 solutions.

If a $\theta$ can be produced by both of these inequalities, then
\[ \theta = \dfrac{(2n + 1) 90^\circ}{420} = \dfrac{m \cdot 180^\circ}{300}, \]or $5(2n + 1) = 14m.$ There are no integer solutions to this equation since the left hand side would be odd while the right hand side would be even.

So we have not overcounted, and there are a total of $840 + 600 = \boxed{1440}$ solutions.